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Wewaii [24]
1 year ago
12

Morgan uses 1/4 of her supply of raisins to make trail mix and 3/8 of her supply of raisins to make cookies. If Morgan uses 5 po

unds of raisins, how many pounds of raisins are in her supply?
Mathematics
1 answer:
Mnenie [13.5K]1 year ago
5 0
First, we have to find how much supply of raisins she uses. It is 1/4+3/8=5/8. It makes the total 5 pounds of raisins that she uses. Constructing an equation, we can find how much raisins are available in her supply. \frac{5}{8} x = 5. And the x= 8. 
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Answer:

30 divided by 1.8= 16.6

Step explanation:

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1 year ago
Lily sold 18 items at the street fair. She sold bracelets for $6 each and necklaces for $5 each for a total of $101. Which syste
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B + N = 18 and 6B + 5N = 101. This is the system of equations you would use.

7 0
2 years ago
Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

x^{2} - 24 = 1

Given that resulting expression is a second order polynomial of the form x^{2} - a^{2}, there are two real and distinct solutions. Roots of the expression are:

x_{1} = -5 and x_{2} = 5.

Now, it is also required to determine which part of the interval (x_{1}, x_{2}) is equal to a number greater than zero (positive). That is:

x^{2} - 24 > 0

x^{2} > 24

x < -4.899 and x > 4.899.

Therefore, exists two sub-intervals: [-5, -4.899] and \left[4.899,5\right]. Besides, x^{2} - 24 > y = 1 in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx

A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}

A = 2.525 -2.474+9.798 + 2.525 - 2.474

A = 9.9\,units^{2}

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0
1 year ago
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(24, 12) and (36, 0).  The least amount of flowering plants occurs when x=2y, and the largest amount occurs when y=0.  These two points satisfy both conditions and both sum to 36.
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Answer:

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Step-by-step explanation:

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